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函数返回引用解决方法

WBOY
Release： 2016-06-13 12:20:21
Original
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函数返回引用
function &bar()
{
    $a = 5;
    return $a;
}
$c=bar();
echo $c;
上面的函数不是运行完毕 函数体的变量不是给释放了 怎么返回的引用还有效呢？
------解决思路----------------------
运行结果不是5么？你return 返回给$c了

Related labels：
bar echo function nbsp return
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